Vinooozz wrote:Dear friends,
I have a permutation sum copied below. I am not able to solve the iii)
Please help me resolve this with explanation
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S, (ii) vowels are all together,
(iii) there are always 4 letters between P and S?
Thanks & Regards,
Vinu
The word PERMUTATIONS has 12 letters to it not all distinct, it has 2 Ts to it as well.
(i) PERMUTATIONS itself is a word that starts with P and ends with S. The remaining 10 letters need arrangements only, and 10 letters with only T occurring twice may be arranged in [spoiler]
10! /2![/spoiler] number of ways. This is cumbersome to solve hence GMAT won't expect you to solve it fully.
(ii) Since PERMUTATIONS has E, U, A, I, and O all 5 vowels to it, and for those vowels to occur together we first need to take them out from PERMUTATIONS, and put them together in a bracket like (EUAIO) and consider this bracket as a single element now so that we now need to permute PRMTTNS(EUAIO) i.e. 8 elements only with only T occurring twice; never forget that the elements in the bracket (EUAIO) can further be permuted within themselves in 5! number of ways. Hence, the arrangements when vowels are all together are [spoiler]
(8! × 5!)/ 2![/spoiler], which again is cumbersome to solve hence GMAT won't expect you to solve it fully.
(iii) Consider
PERMU
STATION as one case of this kind and
SERMU
PTATION as another where there are always 4 letters between P and S. We have 7 allocations possible to fix P and S in that way
PERMU
STATION, and 7 allocations possible to fix S and P in that way
SERMU
PTATION. So, basically there are 14 ways to fix P and S, whereas the remaining 10 letters with only T occurring twice may be arranged in 10! /2! number of ways. Hence, the total arrangements in which there are always 4 letters between P and S is [spoiler]
14 × 10! /2![/spoiler].
